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Question

Quantitative Ability and Data Interpretation Question on SI & CI

A man lends some money to his friend at 5% per annum of interest rate. After 2 years, the difference between the Simple and the compound interest on money is Rs. 50. What will be the value of the amount at the end of 3 years if compounded annually?

A

21325.6

B

24512.5

C

22252.7

D

23152.5

Answer

23152.5

Explanation

Solution

Let's solve the problem step by step.
Given:
- Interest rate r=5%r = 5\% per annum
- The difference between Simple Interest (SI) and Compound Interest (CI) after 2 years is Rs. 50
First, let's denote the principal amount as PP.
Simple Interest (SI) for 2 years:
SI=P×r100×tSI = P \times \frac{r}{100} \times t
For 2 years, t=2t = 2:
SI=P×5100×2=0.1PSI = P \times \frac{5}{100} \times 2 = 0.1P
Compound Interest (CI) for 2 years:
CI=P(1+r100)tPCI = P \left(1 + \frac{r}{100}\right)^t - P
For 2 years, t=2t = 2:
CI=P(1+5100)2P=P(1.052)P=P(1.1025)P=0.1025PCI = P \left(1 + \frac{5}{100}\right)^2 - P = P \left(1.05^2\right) - P = P \left(1.1025\right) - P = 0.1025P
Given the difference between CI and SI after 2 years is Rs. 50:
CISI=0.1025P0.1P=0.0025P=50CI - SI = 0.1025P - 0.1P = 0.0025P = 50
P=500.0025=20000P = \frac{50}{0.0025} = 20000
Now, we need to find the amount at the end of 3 years when compounded annually:
A=P(1+r100)tA = P \left(1 + \frac{r}{100}\right)^t
For 3 years, t=3t = 3:
A=20000(1+5100)3=20000(1.05)3A = 20000 \left(1 + \frac{5}{100}\right)^3 = 20000 \left(1.05\right)^3
Calculating 1.0531.05^3:
1.053=1.1576251.05^3 = 1.157625
A=20000×1.157625=23152.5A = 20000 \times 1.157625 = 23152.5
So, the value of the amount at the end of 3 years if compounded annually is:
Answer: D (23152.5)